I have a spring boot jar file and inside it a manifest file as below
Manifest-Version: 1.0
Implementation-Title: myApp
Implementation-Version: 0.1
Built-By: me
Implementation-Vendor-Id: com.myApp
Spring-Boot-Version: 2.0.0.RELEASE
Main-Class: org.springframework.boot.loader.JarLauncher
Start-Class: com.myApp.smartlight.BootMongoDBApp
Spring-Boot-Classes: BOOT-INF/classes/
Spring-Boot-Lib: BOOT-INF/lib/
Created-By: Apache Maven 3.6.1
Build-Jdk: 1.8.0_151
And inside it is a class file named com.myApp.initiate.Initiator
(packaged under BOOT-INF/classes folder inside the jar) . I am trying to run Initiator class from command line on Windows machine as below
java -cp myApp.jar com.myApp.initiate.Initiator
but no luck. I also tried mentioning classpath in the above command as
java -cp "myApp.jar;BOOT-INF/*" com.myApp.initiate.Initiator
but it still doesn't work.
What am I doing wrong?
Initiator.java
package com.myApp.initiate.Initiator;
public class Initiator {
public static void main(String... args) {
System.out.print("hello");
}
}
Update:
Initiator
class was packaged under BOOT-INF/classes folder. When I copied it at the jar root and tried below command it worked
myApp.jar
|
|--org
|--BOOT-INF
|--META-INF
|--Initiator.class
java -cp myApp.jar Initiator
hello
Standard way to run a Spring Boot app jar is with
java -jar myApp.jar
To run a different class you need to switch to using the org.springframework.boot.loader.PropertiesLauncher
and add the loader.main
property to you command line to choose a different class.
So you manifest must contain:
Main-Class: org.springframework.boot.loader.PropertiesLauncher
Then on the command line:
java -jar myApp.jar -Dloader.main=com.myApp.initiate.Initiator
Have a look at the docs as well for more information: https://docs.spring.io/spring-boot/docs/current/reference/html/appendix-executable-jar-format.html#executable-jar-property-launcher-features
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